Answer
See below.
Work Step by Step
We know that $\log_a x+\log_a y=\log_a (x\cdot y)$, hence here $\log_2 (3x-2)+\log_2 (x)=\log_a (x\cdot (3x-2))$
We know that if $a^x=y$, then $\log_a y=x$ and vice versa.
Hence here $2^4=x(3x-2)\\16=3x^2-2x\\3x^2-2x-16=0\\(x+2)(3x-8)=0$
Thus $x=-2$ (but $x$ must be positive by the definition of logarithms or $x=8/3$. SO $x=8/3$.
